# Complicate density problem

1. Dec 10, 2009

1. The problem statement, all variables and given/known data

We are lowering two small balls into the container filled with oil. We are measuring the final velocities of these balls achieved. For the steel ball we measure the final velocity of v1 = 0.23m/s and for the aluminum ball we measure v2 = 0.06m/ s. What are the density and the viscosity of oil, if the radius's of both balls are r = 3mm; the density of steel is ρ1 = 7800 kg/m3, and the density of aluminum is ρ2 = 2700 kg/m3. Assume that the linear law of resistance applies.

2. Relevant equations

ρ= m/V

3. The attempt at a solution

Volume of the sphere= 4/3 πr³
1000 kg/m³= 1 g/cm³
r= 3 mm= 0.3 cm
ρ= 7800 kg/m³= 7.8 g/cm³
ρ= 2700 kg/m³= 2.7 g/cm³

V= 0.11304 cm³

ρ = m/V → m= ρV

m(steal ball)= 0.88 g
m(aluminum ball)= 0.31 g

And this is all I managed to calculate. I know that conservation of mass and conservation of momentum apply to the density problems, but I don’t know how to apply them to find the density of oil.
Thank you!

2. Dec 10, 2009

### Staff: Mentor

3. Dec 10, 2009

Calculation for the steal ball:

Fd= Fg
6πμrv(1)= m(1)g
μ= m(1)g/ 6πrv(1)
μ= 666.81 Pa s

Calculation for the aluminum ball:

Fd= Fg
6πμrv(2)= m(2)g
μ= m(2)g/ 6πrv(2)
μ= 1045.54 Pa s

So I calculated the viscosity according to Stokes' equation, however I had two balls in the question, one of steel and the other of aluminum so I just went ahead and plugged in both of them in seperate equations, and that gave me two different vicosities.

How do I continue?

4. Dec 10, 2009

### Staff: Mentor

You can't calculate viscosity not knowing density.

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5. Dec 10, 2009

Can you please, give me another hint for how to approach my problem?
I'm lost and I would really like to understand the problem and find the solution.

6. Dec 10, 2009

### trust143_raj

You use the Stoke law n(viscosity)=[2(density of steel-density of oil)*g*R]/(9v1)
v1 is the velocity for steel
Again use the same formula for The case of aluminium and divide one equation with other..so that only the unknown will be the density of oil

7. Dec 10, 2009

### Staff: Mentor

In other words - you have two equations in two unknowns.

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methods

8. Dec 10, 2009

OK, I got the final formula:

ρ (oil)= [ρ(steal)v(aluminum) – ρ(aluminum)v(steal)] / [v(aluminum) – v(steal)]
ρ(oil)= 900 kg/m³

And now for viscosity:

μ= 2/9 [ρ(steal) – ρ(oil) / v(steal)] gr²
μ= 5.88 Pa s

Hopefully, my calculations are now correct!?:shy:
Thank you for helping!